Question

$\Delta H$ and $\Delta S$ for the reaction, $2A+B \rightarrow C$ at 298 K are $400~kJ~mol^{-1}$ and $2~kJ~K^{-1}mol^{-1}$ respectively. At or above $T(K)$, the reaction becomes spontaneous. What is $T(K)$?

• A. 100
• B. 200
• C. 150
• D. 125 \bigskip

Hint:

For reactions with \[ \Delta H>0,\qquad \Delta S>0 \] high temperature favors spontaneity. The critical temperature is \[ T=\frac{\Delta H}{\Delta S} \]

Answer 1

The Correct Option is B

Concept: The spontaneity of a reaction is determined by Gibbs free energy. \[ \Delta G=\Delta H-T\Delta S \] A reaction becomes spontaneous when \[ \Delta G<0 \] The limiting condition occurs when \[ \Delta G=0 \] which gives the transition temperature.

Step 1: Write the Gibbs free energy equation.
\[ \Delta G=\Delta H-T\Delta S \] Given \[ \Delta H=400~kJ~mol^{-1} \] \[ \Delta S=2~kJ~K^{-1}mol^{-1} \]

Step 2: Find the temperature at which the reaction just becomes spontaneous.
At equilibrium boundary, \[ \Delta G=0 \] Therefore, \[ 0=\Delta H-T\Delta S \] \[ T=\frac{\Delta H}{\Delta S} \] Substituting values, \[ T=\frac{400}{2} \] \[ T=200~K \]

Step 3: Interpret the result.
For \[ T>200K \] \[ \Delta G<0 \] Hence the reaction becomes spontaneous at or above \[ \boxed{200K} \]